If $ - \frac{\pi }{2} < x < \frac{\pi }{2}$, then the value of $\log \,\,\sec \,x$ is

$2\,\,{\coth ^{ - 1}}\left( {{\rm{cose}}{{\rm{c}}^2}\,\frac{x}{2} - 1} \right)$

$2\,\,{\coth ^{ - 1}}\left( {{\rm{cose}}{{\rm{c}}^2}\frac{x}{2} + 1} \right)$

$2\,\,{\rm{cosec}}{{\rm{h}}^{ - 1}}\left( {{\rm{co}}{{\rm{t}}^2}\frac{x}{2} - 1} \right)$

$2\,{\rm{cosec}}{{\rm{h}}^{ - 1}}\left( {{\rm{co}}{{\rm{t}}^2}\frac{x}{2} + 1} \right)$

${\rm{sec}}{{\rm{h}}^{ - 1}}\,\,\left( {\frac{1}{2}} \right)$ is

$\log \,\,\left( {\sqrt 3 + \sqrt 2 } \right)$

$\log \,\,\left( {\sqrt 3 + 1} \right)$

${\rm{sec}}{{\rm{h}}^{ - 1}}\,\,\left( {\sin \,\,\theta } \right)$ is equal to

$\log \,\,\tan \,\,\frac{\theta }{2}$

$\log \,\,\sin \,\,\frac{\theta }{2}$

$\log \,\,\cos \,\,\frac{\theta }{2}$

The real part of $\sin \log (x+i y)$ is

$\sin \left(\log \sqrt{x^2+y^2}\right) \sinh \left(\tan ^{-1} \frac{y}{x}\right)$

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JEE • MATHEMATICS • HYPERBOLIC FUNCTIONS

Hyperbolic functions practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Easy

1 of 15

Application +4 | -1

\cos (x+h y)

is cqual to:

View answer and explanation

Correct option: C

\cos x \cosh y-i \sin x \sinh y

Option C (cos x cosh y - i sin x sinh y) is correct as it represents the expansion of cos(x + iy) using Euler's formula and hyperbolic functions.

2 Easy

2 of 15

Application +4 | -1

The imaginary part of

\log \sin (x+i y)

is:

View answer and explanation

Correct option: B

\tan ^{-1}\left(\frac{\cos x \sinh y}{\sin x \cosh y}\right)

Option B (

\tan ^{-1}(\frac{{\cos x \sinh y}}{{\sin x \cosh y}})

) is correct because it correctly identifies the imaginary part of the expression after applying trigonometric identities and logarithmic properties.

3 Easy

3 of 15

Application +4 | -1

\frac{{1 + \mathrm{tanh\;}\,\,\frac{x}{2}}}{{1 - \mathrm{tanh\;}\,\,\frac{x}{2}}}

is equal to

View answer and explanation

Correct option: B

{e^x}

Option B (

{e^x}

) is correct because simplifying the hyperbolic tangent expression using exponential forms results in

{e^x}

4 Easy

4 of 15

Understanding +4 | -1

Which of the following functions is not defined at

x=0

View answer and explanation

Correct option: B

\operatorname{cosech} x

Option B (

\operatorname{cosech} x

) is correct because

\operatorname{cosech} x = \frac{1}{\sinh x}

and

\sinh x = 0

x=0

, making

\operatorname{cosech} x

undefined at

x=0

5 Easy

5 of 15

Application +4 | -1

The real and imaginary part of

2 \sin (\alpha+i \beta) \cos (\alpha-i \beta)

are

View answer and explanation

Correct option: C

\sin 2 \alpha, \sinh 2 \beta

Option C (

\sin 2 \alpha, \sinh 2 \beta

) is correct because it correctly identifies the real and imaginary parts after applying trigonometric identities and hyperbolic functions.

6 Easy

6 of 15

Application +4 | -1

- \frac{\pi }{2} < x < \frac{\pi }{2}

, then the value of

\log \,\,\sec \,x

View answer and explanation

Correct option: A

2\,\,{\coth ^{ - 1}}\left( {\mathrm{cose}{\mathrm{c}^2}\,\frac{x}{2} - 1} \right)

Option A (

2\,\,{\coth ^{ - 1}}\left( {\mathrm{cose}{\mathrm{c}^2}\,\frac{x}{2} - 1} \right)

) is correct because it is the result of applying componendo and dividendo rule and hyperbolic identities to the given logarithmic expression.

7 Easy

7 of 15

Understanding +4 | -1

\frac{1+\tanh x}{1-\tanh x}

is equal to:

View answer and explanation

Correct option: A

e^{2 x}

Option A (

e^{2 x}

) is correct because by converting

\tanh x

to its exponential form and simplifying the expression, it reduces to

e^{2x}

8 Easy

8 of 15

Application +4 | -1

{\cosh ^{ - 1}}\,\,x = \log \,\,\left( {2 + \sqrt 3 } \right)

, then

x

is equal to

View answer and explanation

Correct option: A

Option A is correct because it uses the inverse hyperbolic cosine definition to find the value of x.

9 Easy

9 of 15

Application +4 | -1

\mathrm{sec}{\mathrm{h}^{ - 1}}\,\,\left( {\frac{1}{2}} \right)

View answer and explanation

Correct option: C

\log \,\,\left( {2 + \sqrt 3 } \right)

Option C (

\log \,\left( {2 + \sqrt 3 } \right)

) is correct because it is the result of applying the inverse hyperbolic function formula for sech^-1(x).

10 Easy

10 of 15

Application +4 | -1

\mathrm{sec}{\mathrm{h}^{ - 1}}\,\,\left( {\sin \,\,\theta } \right)

is equal to

View answer and explanation

Correct option: D

\log \,\,\cot \,\,\frac{\theta }{2}

Option D (

\log \,\cot \,\frac{\theta }{2}

) is correct because the inverse hyperbolic secant simplifies to this logarithmic form using hyperbolic identities.

11 Easy

11 of 15

Application +4 | -1

The imaginary part of

\cosh (\alpha+1 \beta)

is:

View answer and explanation

Correct option: B

\sinh \alpha \sin \beta

Option B (

\sinh \alpha \sin \beta

) is correct because it is the imaginary part of the expansion of

\cosh (\alpha+i \beta)

using hyperbolic trigonometric identities.

12 Easy

12 of 15

Application +4 | -1

2\,\,{\tanh ^{ - 1}}\,\frac{1}{2}

is equal to

View answer and explanation

Correct option: C

\log \,\,3

Option C (

\log \,3

) is correct because using the identity for inverse hyperbolic tangent and simplifying leads to

\log \,3

13 Easy

13 of 15

Application +4 | -1

The imaginary part of

\sin ^2(x+i y)

is:

View answer and explanation

Correct option: D

\frac{1}{2} \sin 2 x \sinh 2 y

Option D (

\frac{{1}}{{2}} \sin 2 x \sinh 2 y

) is correct because it correctly identifies the imaginary part of the expression after applying trigonometric identities and hyperbolic functions.

14 Easy

14 of 15

Application +4 | -1

\frac{\operatorname{cosech} x}{\sqrt{\operatorname{cosech}^2 x+1}}

equals:

View answer and explanation

Correct option: C

\operatorname{sech} x

Option C (sech x) is correct because it is the simplified form of the given expression using hyperbolic identities.

15 Easy

15 of 15

Application +4 | -1

The real part of

\sin \log (x+i y)

View answer and explanation

Correct option: C

\sin \left(\log \sqrt{x^2+y^2}\right) \cosh \left(\tan ^{-1} \frac{y}{x}\right)

Option C (

\sin (\log \sqrt{{x^2+y^2}}) \cosh (\tan ^{-1} \frac{{y}}{{x}})

) is correct because it correctly identifies the real part of the expression after applying trigonometric identities and logarithmic properties.

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Key Takeaways

✓ This set contains 15 single-correct-answer MCQs on HYPERBOLIC FUNCTIONS.
✓ Try each question before opening its answer. Read the explanation where one is available.
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